Self Ionization of Water and Buffer Class 11 Notes | EduRev

Chemistry for JEE

Class 11 : Self Ionization of Water and Buffer Class 11 Notes | EduRev

 Page 1


 
   
        
Self ionization of water
Concentration of H
2
O ([H
2
O]) = 
1000 / 18
  55.5 M
1
?
Water is very weak electrolyte
H
2
O      H
+
   +   OH
–
55.5
55.5 – 10
–7
      10
–7
        10
–7
   (at 25
o
C)
?
O] [H
] [OH ] [H
k
2
–
eq
?
?
            Constant
? K
w
 = k
eq
 [H
2
O]  = [H
+
] [OH
–
]
? K
w
 = [H
+
]  [OH
–
],at  25
o
C, [H
+
] = [OH
–
] = 10
–7
K
w
 (25
o
C)    = 10
–14
   
?
Self ionization constant of water
let,
1
w
K
 at temperature  T
1
  ,
2
w
K
 at temperature  T
2
?
?
?
?
?
?
? ?
?
2 1 w
w
T
1
 – 
T
1
 
R 2.303
H
  
K
K
  log
1
2
, ? ? ? K      T
w
? For calculation of ? :
H
2
O      H
+
  +  OH
–
 c         0 0
 c  (1 – ?)       c ? ? ??? ? ? ?? ?c ?
Page 2


 
   
        
Self ionization of water
Concentration of H
2
O ([H
2
O]) = 
1000 / 18
  55.5 M
1
?
Water is very weak electrolyte
H
2
O      H
+
   +   OH
–
55.5
55.5 – 10
–7
      10
–7
        10
–7
   (at 25
o
C)
?
O] [H
] [OH ] [H
k
2
–
eq
?
?
            Constant
? K
w
 = k
eq
 [H
2
O]  = [H
+
] [OH
–
]
? K
w
 = [H
+
]  [OH
–
],at  25
o
C, [H
+
] = [OH
–
] = 10
–7
K
w
 (25
o
C)    = 10
–14
   
?
Self ionization constant of water
let,
1
w
K
 at temperature  T
1
  ,
2
w
K
 at temperature  T
2
?
?
?
?
?
?
? ?
?
2 1 w
w
T
1
 – 
T
1
 
R 2.303
H
  
K
K
  log
1
2
, ? ? ? K      T
w
? For calculation of ? :
H
2
O      H
+
  +  OH
–
 c         0 0
 c  (1 – ?)       c ? ? ??? ? ? ?? ?c ?
  
 
K
w
 = c ? × c ?
?
c
k
    
w
? ? c = 5 5.56 M ?
w
–2
K 10  1.8  ? ? ?
?
pH
potent 
    ?
strength
H ydrogen ion
(pH s cale is given by S orension)
? pH     =  – log [H
+
]
p OH =  – log [OH
–
]
p K
w
  =  – log K
w
 
at 25°C
Acidic basic
0
7 14
Neutral
p scale
H
at any temperature :
[H
+
] = [OH
–
]
[H
+
]  [OH
–
] = K
w
? – log [H
+
] – log [OH
–
] = – log [K
w
]
? pH + pOH = p K
w
at 25
o
C at 80
o
C
K
w
 = 10
–14
K
w
 = 10
–12
pH + pOH = 14 pH + pOH = 12
Some concept regarding pH calculation
? Concept 1 : ? Concept 2:
   5 L
pH = 3  ?
  1 L
pH = ?       
   3 L
pH = 3  + 
   2 L
pH = 3  ? 
   5 L
pH = ?
? Concept 3: ? Concept 4:
   2 L
pH = 3 +
   8 L
   HO
2
=
  10 L
pH = ?   
   2 L
pH = 13 +
   8 L
 water =
  10 L
pH = ?
? Note :
7
pH
(i) Calculate pOH
(ii) pH = 14- pOH
Calculate  pH
? Concept 5:
  pH = 3
 +
  pH = 4
 = 
  pH = ?
Page 3


 
   
        
Self ionization of water
Concentration of H
2
O ([H
2
O]) = 
1000 / 18
  55.5 M
1
?
Water is very weak electrolyte
H
2
O      H
+
   +   OH
–
55.5
55.5 – 10
–7
      10
–7
        10
–7
   (at 25
o
C)
?
O] [H
] [OH ] [H
k
2
–
eq
?
?
            Constant
? K
w
 = k
eq
 [H
2
O]  = [H
+
] [OH
–
]
? K
w
 = [H
+
]  [OH
–
],at  25
o
C, [H
+
] = [OH
–
] = 10
–7
K
w
 (25
o
C)    = 10
–14
   
?
Self ionization constant of water
let,
1
w
K
 at temperature  T
1
  ,
2
w
K
 at temperature  T
2
?
?
?
?
?
?
? ?
?
2 1 w
w
T
1
 – 
T
1
 
R 2.303
H
  
K
K
  log
1
2
, ? ? ? K      T
w
? For calculation of ? :
H
2
O      H
+
  +  OH
–
 c         0 0
 c  (1 – ?)       c ? ? ??? ? ? ?? ?c ?
  
 
K
w
 = c ? × c ?
?
c
k
    
w
? ? c = 5 5.56 M ?
w
–2
K 10  1.8  ? ? ?
?
pH
potent 
    ?
strength
H ydrogen ion
(pH s cale is given by S orension)
? pH     =  – log [H
+
]
p OH =  – log [OH
–
]
p K
w
  =  – log K
w
 
at 25°C
Acidic basic
0
7 14
Neutral
p scale
H
at any temperature :
[H
+
] = [OH
–
]
[H
+
]  [OH
–
] = K
w
? – log [H
+
] – log [OH
–
] = – log [K
w
]
? pH + pOH = p K
w
at 25
o
C at 80
o
C
K
w
 = 10
–14
K
w
 = 10
–12
pH + pOH = 14 pH + pOH = 12
Some concept regarding pH calculation
? Concept 1 : ? Concept 2:
   5 L
pH = 3  ?
  1 L
pH = ?       
   3 L
pH = 3  + 
   2 L
pH = 3  ? 
   5 L
pH = ?
? Concept 3: ? Concept 4:
   2 L
pH = 3 +
   8 L
   HO
2
=
  10 L
pH = ?   
   2 L
pH = 13 +
   8 L
 water =
  10 L
pH = ?
? Note :
7
pH
(i) Calculate pOH
(ii) pH = 14- pOH
Calculate  pH
? Concept 5:
  pH = 3
 +
  pH = 4
 = 
  pH = ?
 
2       :     1
? Note : The final pH of solution after mixing two solution is in between the previous solution pH.
? Concept 6: ? Concept 7:
  pH = 3
V
 +   pH = 11
V
?   pH = ?
2 V
  pH = ?
10 M HCl
–8
? Note :
When [H
+
] > 10
–5
 ? [H
+
]
water
  neglected
When [H
+
] < 10
–5
 ? [H
+
]
water
  considered
H
2
O      H
+
    +   OH
–
55.56         10
–8
55.56 – x        10
–8
 + x    x
? x (10
–8 
+ x) = 10
–14      
       ?      x = .94 × 10
–7
? [H
+
] = 10
–8
 + .94 × 10
–7
= 1.04 × 10
–7
? pH = 7 – log (1.04)
? Concept 8:
  pH = ?
10 M H Cl
2
? Dissociation of weak acid :
H
+ 
 A
–
       H
+
  +  A
–
c
c (1– ?) c ?      c ?
?
?
?
– 1
c
  K
2
a
for weak acid ? ?< < 1 K
a
 =  c ?
2
[H
+
] = c ? ? = 
a
K c , pH = – log 
a
K c ,   for  weak bases    [H
+
] = 
b
w
K c
K
?
 =
] OH [
K
w
?
? Note :
(a) HA
1
c
K
a
1
||
HA
2
c
K
a
2
?
2
1
a
a
2
1
K
K
  
] H [
] H [
?
?
?
Page 4


 
   
        
Self ionization of water
Concentration of H
2
O ([H
2
O]) = 
1000 / 18
  55.5 M
1
?
Water is very weak electrolyte
H
2
O      H
+
   +   OH
–
55.5
55.5 – 10
–7
      10
–7
        10
–7
   (at 25
o
C)
?
O] [H
] [OH ] [H
k
2
–
eq
?
?
            Constant
? K
w
 = k
eq
 [H
2
O]  = [H
+
] [OH
–
]
? K
w
 = [H
+
]  [OH
–
],at  25
o
C, [H
+
] = [OH
–
] = 10
–7
K
w
 (25
o
C)    = 10
–14
   
?
Self ionization constant of water
let,
1
w
K
 at temperature  T
1
  ,
2
w
K
 at temperature  T
2
?
?
?
?
?
?
? ?
?
2 1 w
w
T
1
 – 
T
1
 
R 2.303
H
  
K
K
  log
1
2
, ? ? ? K      T
w
? For calculation of ? :
H
2
O      H
+
  +  OH
–
 c         0 0
 c  (1 – ?)       c ? ? ??? ? ? ?? ?c ?
  
 
K
w
 = c ? × c ?
?
c
k
    
w
? ? c = 5 5.56 M ?
w
–2
K 10  1.8  ? ? ?
?
pH
potent 
    ?
strength
H ydrogen ion
(pH s cale is given by S orension)
? pH     =  – log [H
+
]
p OH =  – log [OH
–
]
p K
w
  =  – log K
w
 
at 25°C
Acidic basic
0
7 14
Neutral
p scale
H
at any temperature :
[H
+
] = [OH
–
]
[H
+
]  [OH
–
] = K
w
? – log [H
+
] – log [OH
–
] = – log [K
w
]
? pH + pOH = p K
w
at 25
o
C at 80
o
C
K
w
 = 10
–14
K
w
 = 10
–12
pH + pOH = 14 pH + pOH = 12
Some concept regarding pH calculation
? Concept 1 : ? Concept 2:
   5 L
pH = 3  ?
  1 L
pH = ?       
   3 L
pH = 3  + 
   2 L
pH = 3  ? 
   5 L
pH = ?
? Concept 3: ? Concept 4:
   2 L
pH = 3 +
   8 L
   HO
2
=
  10 L
pH = ?   
   2 L
pH = 13 +
   8 L
 water =
  10 L
pH = ?
? Note :
7
pH
(i) Calculate pOH
(ii) pH = 14- pOH
Calculate  pH
? Concept 5:
  pH = 3
 +
  pH = 4
 = 
  pH = ?
 
2       :     1
? Note : The final pH of solution after mixing two solution is in between the previous solution pH.
? Concept 6: ? Concept 7:
  pH = 3
V
 +   pH = 11
V
?   pH = ?
2 V
  pH = ?
10 M HCl
–8
? Note :
When [H
+
] > 10
–5
 ? [H
+
]
water
  neglected
When [H
+
] < 10
–5
 ? [H
+
]
water
  considered
H
2
O      H
+
    +   OH
–
55.56         10
–8
55.56 – x        10
–8
 + x    x
? x (10
–8 
+ x) = 10
–14      
       ?      x = .94 × 10
–7
? [H
+
] = 10
–8
 + .94 × 10
–7
= 1.04 × 10
–7
? pH = 7 – log (1.04)
? Concept 8:
  pH = ?
10 M H Cl
2
? Dissociation of weak acid :
H
+ 
 A
–
       H
+
  +  A
–
c
c (1– ?) c ?      c ?
?
?
?
– 1
c
  K
2
a
for weak acid ? ?< < 1 K
a
 =  c ?
2
[H
+
] = c ? ? = 
a
K c , pH = – log 
a
K c ,   for  weak bases    [H
+
] = 
b
w
K c
K
?
 =
] OH [
K
w
?
? Note :
(a) HA
1
c
K
a
1
||
HA
2
c
K
a
2
?
2
1
a
a
2
1
K
K
  
] H [
] H [
?
?
?
  
 
? So HA
2
 is stronger acid than HA
1
(b) HA
1
c
1
K
a
1
HA
2
  c
2
K
a
2
2 2
1 1
2
1
Ka c
ka c
    
] H [
] H [
?
?
?
? The acidic strength of two weak acid can be compared by K
a
 value only in the case when their concen-
tration are same. Otherwise the procedure of case  (b) should be followed .
?
pH  of mixture of two acids
?
? Case (I) (Strong acid + weak acid)
Question : HCl (0.1 M) + CH
3
COOH (0.2 M)
  
?
K
a
 = 10
–5
CH
3
COOH  CH
3
COO
–
   +   H
+
t = 0     0.2      0          0. 1
t = t
eq
     0.2 – x      x          0.1 + x (x is very small)
      
?
      
?
  
?
     0.2       x 0.1
? pH = – log (0.1) = 1 ? [CH
3
COO
–
] = ? ?
5 – –
3
5 –
10  2  ] COO [CH    
2 . 0
1 . 0 x
  10 ? ? ?
?
?
?    Case (II-A) (weak acid + weak acid ) (k
as
 are incomparable)
Question : HA
1
 (0.1 M,  
K
a
1
 = 10
–5
)
HA
2
 (0.2 M,  
K
a
2
 = 10
–9
)
HA
1     
    H
+
   + A
1
–
HA
2     
    H
+
   + A
2
–
0.1      0 0 0.2        0 0
0.1 – x     x + y x 0.2 – y        x + y  y       (
K
a
1
 >> 
K
a
2
?
       
? ? ?
          
?
 
?
     ? x  > > y)
0.1       x x  0.2           x  y
x and y are also very less
Page 5


 
   
        
Self ionization of water
Concentration of H
2
O ([H
2
O]) = 
1000 / 18
  55.5 M
1
?
Water is very weak electrolyte
H
2
O      H
+
   +   OH
–
55.5
55.5 – 10
–7
      10
–7
        10
–7
   (at 25
o
C)
?
O] [H
] [OH ] [H
k
2
–
eq
?
?
            Constant
? K
w
 = k
eq
 [H
2
O]  = [H
+
] [OH
–
]
? K
w
 = [H
+
]  [OH
–
],at  25
o
C, [H
+
] = [OH
–
] = 10
–7
K
w
 (25
o
C)    = 10
–14
   
?
Self ionization constant of water
let,
1
w
K
 at temperature  T
1
  ,
2
w
K
 at temperature  T
2
?
?
?
?
?
?
? ?
?
2 1 w
w
T
1
 – 
T
1
 
R 2.303
H
  
K
K
  log
1
2
, ? ? ? K      T
w
? For calculation of ? :
H
2
O      H
+
  +  OH
–
 c         0 0
 c  (1 – ?)       c ? ? ??? ? ? ?? ?c ?
  
 
K
w
 = c ? × c ?
?
c
k
    
w
? ? c = 5 5.56 M ?
w
–2
K 10  1.8  ? ? ?
?
pH
potent 
    ?
strength
H ydrogen ion
(pH s cale is given by S orension)
? pH     =  – log [H
+
]
p OH =  – log [OH
–
]
p K
w
  =  – log K
w
 
at 25°C
Acidic basic
0
7 14
Neutral
p scale
H
at any temperature :
[H
+
] = [OH
–
]
[H
+
]  [OH
–
] = K
w
? – log [H
+
] – log [OH
–
] = – log [K
w
]
? pH + pOH = p K
w
at 25
o
C at 80
o
C
K
w
 = 10
–14
K
w
 = 10
–12
pH + pOH = 14 pH + pOH = 12
Some concept regarding pH calculation
? Concept 1 : ? Concept 2:
   5 L
pH = 3  ?
  1 L
pH = ?       
   3 L
pH = 3  + 
   2 L
pH = 3  ? 
   5 L
pH = ?
? Concept 3: ? Concept 4:
   2 L
pH = 3 +
   8 L
   HO
2
=
  10 L
pH = ?   
   2 L
pH = 13 +
   8 L
 water =
  10 L
pH = ?
? Note :
7
pH
(i) Calculate pOH
(ii) pH = 14- pOH
Calculate  pH
? Concept 5:
  pH = 3
 +
  pH = 4
 = 
  pH = ?
 
2       :     1
? Note : The final pH of solution after mixing two solution is in between the previous solution pH.
? Concept 6: ? Concept 7:
  pH = 3
V
 +   pH = 11
V
?   pH = ?
2 V
  pH = ?
10 M HCl
–8
? Note :
When [H
+
] > 10
–5
 ? [H
+
]
water
  neglected
When [H
+
] < 10
–5
 ? [H
+
]
water
  considered
H
2
O      H
+
    +   OH
–
55.56         10
–8
55.56 – x        10
–8
 + x    x
? x (10
–8 
+ x) = 10
–14      
       ?      x = .94 × 10
–7
? [H
+
] = 10
–8
 + .94 × 10
–7
= 1.04 × 10
–7
? pH = 7 – log (1.04)
? Concept 8:
  pH = ?
10 M H Cl
2
? Dissociation of weak acid :
H
+ 
 A
–
       H
+
  +  A
–
c
c (1– ?) c ?      c ?
?
?
?
– 1
c
  K
2
a
for weak acid ? ?< < 1 K
a
 =  c ?
2
[H
+
] = c ? ? = 
a
K c , pH = – log 
a
K c ,   for  weak bases    [H
+
] = 
b
w
K c
K
?
 =
] OH [
K
w
?
? Note :
(a) HA
1
c
K
a
1
||
HA
2
c
K
a
2
?
2
1
a
a
2
1
K
K
  
] H [
] H [
?
?
?
  
 
? So HA
2
 is stronger acid than HA
1
(b) HA
1
c
1
K
a
1
HA
2
  c
2
K
a
2
2 2
1 1
2
1
Ka c
ka c
    
] H [
] H [
?
?
?
? The acidic strength of two weak acid can be compared by K
a
 value only in the case when their concen-
tration are same. Otherwise the procedure of case  (b) should be followed .
?
pH  of mixture of two acids
?
? Case (I) (Strong acid + weak acid)
Question : HCl (0.1 M) + CH
3
COOH (0.2 M)
  
?
K
a
 = 10
–5
CH
3
COOH  CH
3
COO
–
   +   H
+
t = 0     0.2      0          0. 1
t = t
eq
     0.2 – x      x          0.1 + x (x is very small)
      
?
      
?
  
?
     0.2       x 0.1
? pH = – log (0.1) = 1 ? [CH
3
COO
–
] = ? ?
5 – –
3
5 –
10  2  ] COO [CH    
2 . 0
1 . 0 x
  10 ? ? ?
?
?
?    Case (II-A) (weak acid + weak acid ) (k
as
 are incomparable)
Question : HA
1
 (0.1 M,  
K
a
1
 = 10
–5
)
HA
2
 (0.2 M,  
K
a
2
 = 10
–9
)
HA
1     
    H
+
   + A
1
–
HA
2     
    H
+
   + A
2
–
0.1      0 0 0.2        0 0
0.1 – x     x + y x 0.2 – y        x + y  y       (
K
a
1
 >> 
K
a
2
?
       
? ? ?
          
?
 
?
     ? x  > > y)
0.1       x x  0.2           x  y
x and y are also very less
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1 –
5 –
10
x . x 
  10 ?
[H
+
] = x = 10
–3  
,  ? pH = 3 [A
1
– 
] = [A
2
–
] =
? Case (II-B) (weak acid + weak acid ) (k
as
 are comparable)
?Question : HA
1
 (0.1)
K
a
1
 = 10
–5
HA
2
 (0.2)
K
a
2
 = 10
–6
HA
1     
 H
+
   + A
1
–
HA
2     
    H
+
    + A
2
–
0.1      0 0 0.2        0 0
0.1 – x     x + y x 0.2 – y        x + y y
0.1     x + y x 0.2        x + y y
? now
0.1
 y) (x x 
   K
1
a
?
?
?
1
a
c
y) (x x 
   K
1
?
?
,
2
a
c
y y) (x  
   K
2
?
?
?
?
2 1
a 2 a 1
K c  K  c y x  ] [H ? ? ? ?
?
? weak polyprotic acids or bases
(H
2
S, H
3
PO
4
, NH
2
 – NH
2
, H
2
CO
3
)
(a) H
2
S     H
+
 + HS
–
K
a
1
HS
–
     H
+
 + S
2–
K
a
2
(b) NH
2
 –  NH
2
 + H
+
       NH
2
 – NH
3
+
1
b
K
NH
2 
–  NH
3
+
 + H
+
     
 H N – H N
3 3
? ?
2
b
K
Cases
    
?   Case I : H
2
S (0.1 M
K
a
1
 = 10
–4
 , 
K
a
2
 = 10
–9
)
? H
2
S
     
  H
+
   + H S
–
H S
– 
   
    H
+
    + S
2–
0.1       0 0 x        x 0
0.1 – x     x + y x – y x – y        x + y y
?
       
?
 
? ?
         
?
       
?
0.1        x x x          x y
K
a
1
 > >  
K
a
2
 x    > >  y
? ? ?  x    
0.1
x
   10
2
4 –
5 –
10
9 –
10   
x
y  x
?
?
Read More
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